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\(=\dfrac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x-1}\)
\(=\dfrac{x+1-1}{x-1}=\dfrac{x}{x-1}\)
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
bổ sung \(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\\x\ne2\end{matrix}\right.\)
A=\(\left(\dfrac{1}{x+1}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)
=\(\left(\dfrac{1\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)
=\(\left(\dfrac{x-1}{\left(x+1\right)\left(x-1\right)}-\dfrac{1}{x^2-1}\right).\dfrac{x+1}{x-2}\)
=\(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x+1}{x-2}\)
=\(\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x-2\right)}\)
=\(\dfrac{1}{x-1}\)